Recursively remove all adjacent duplicates

Given a string, recursively remove adjacent duplicate characters from the string. The output string should not have any adjacent duplicates. See following examples.
Examples

Input: azxxzy 
Output: ay 
First “azxxzy” is reduced to “azzy”. 
The string “azzy” contains duplicates, 
so it is further reduced to “ay”.
Input: geeksforgeeg 
Output: gksfor 
First “geeksforgeeg” is reduced to 
“gksforgg”. The string “gksforgg” 
contains duplicates, so it is further 
reduced to “gksfor”.
Input: caaabbbaacdddd 
Output: Empty String
Input: acaaabbbacdddd 
Output: acac 

The following approach can be followed to remove duplicates in O(N) time: 

  • Start from the leftmost character and remove duplicates at left corner if there are any.
  • The first character must be different from its adjacent now. Recur for string of length n-1 (string without first character).
  • Let the string obtained after reducing right substring of length n-1 be rem_str. There are three possible cases 
    1. If first character of rem_str matches with the first character of original string, remove the first character from rem_str.
    2. If remaining string becomes empty and last removed character is same as first character of original string. Return empty string.
    3. Else, append the first character of the original string at the beginning of rem_str.
  • Return rem_str.

Below image is a dry run of the above approach:



Below is the implementation of the above approach:

C++

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// C/C++ program to remove all
// adjacent duplicates from a string
#include <iostream>
#include <string.h>
 
using namespace std;
 
// Recursively removes adjacent
// duplicates from str and returns
// new string. las_removed is a
// pointer to last_removed character
char* removeUtil(char *str, char *last_removed)
{
     
    // If length of string is 1 or 0
    if (str[0] == '\0' || str[1] == '\0')
        return str;
 
    // Remove leftmost same characters
    // and recur for remaining
    // string
    if (str[0] == str[1])
    {
        *last_removed = str[0];
        while (str[1] && str[0] == str[1])
            str++;
        str++;
        return removeUtil(str, last_removed);
    }
 
    // At this point, the first character
    // is definiotely different
    // from its adjacent. Ignore first
    // character and recursively
    // remove characters from remaining string
    char* rem_str = removeUtil(str+1,
                               last_removed);
 
    // Check if the first character
    // of the rem_string matches with
    // the first character of the
    // original string
    if (rem_str[0] && rem_str[0] == str[0])
    {
        *last_removed = str[0];
       
        // Remove first character
        return (rem_str+1);
    }
 
    // If remaining string becomes
    // empty and last removed character
    // is same as first character of
    // original string. This is needed
    // for a string like "acbbcddc"
    if (rem_str[0] == '\0' &&
                 *last_removed == str[0])
        return rem_str;
 
    // If the two first characters
    // of str and rem_str don't match,
    // append first character of str
    // before the first character of
    // rem_str.
    rem_str--;
    rem_str[0] = str[0];
    return rem_str;
}
 
// Function to remove
char *remove(char *str)
{
    char last_removed = '\0';
    return removeUtil(str, &last_removed);
}
 
// Driver program to test
// above functions
int main()
{
    char str1[] = "geeksforgeeg";
    cout << remove(str1) << endl;
 
    char str2[] = "azxxxzy";
    cout << remove(str2) << endl;
 
    char str3[] = "caaabbbaac";
    cout << remove(str3) << endl;
 
    char str4[] = "gghhg";
    cout << remove(str4) << endl;
 
    char str5[] = "aaaacddddcappp";
    cout << remove(str5) << endl;
 
    char str6[] = "aaaaaaaaaa";
    cout << remove(str6) << endl;
 
    char str7[] = "qpaaaaadaaaaadprq";
    cout << remove(str7) << endl;
 
    char str8[] = "acaaabbbacdddd";
    cout << remove(str8) << endl;
 
    char str9[] = "acbbcddc";
    cout << remove(str9) << endl;
 
    return 0;
}

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Java

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// Java program to remove
// all adjacent duplicates
// from a string
import java.io.*;
import java.util.*;
 
class GFG
{
   
  // Recursively removes adjacent
  //  duplicates from str and returns
  // new string. las_removed is a
  // pointer to last_removed character
  static String removeUtil(String str,
                            char last_removed)
  {
     
    // If length of string is 1 or 0
    if (str.length() == 0 || str.length() == 1)
      return str;
 
    // Remove leftmost same characters
    // and recur for remaining 
    // string
    if (str.charAt(0) == str.charAt(1))
    {
      last_removed = str.charAt(0);
      while (str.length() > 1 && str.charAt(0) ==
                                   str.charAt(1))
        str = str.substring(1, str.length());
      str = str.substring(1, str.length());
      return removeUtil(str, last_removed);
    }
 
    // At this point, the first
    // character is definiotely different 
    // from its adjacent. Ignore first
    // character and recursively 
    // remove characters from remaining string
    String rem_str = removeUtil(str.substring(
                  1,str.length()), last_removed);
 
    // Check if the first character of
    // the rem_string matches with 
    // the first character of the original string
    if (rem_str.length() != 0 &&
             rem_str.charAt(0) == str.charAt(0))
    {
      last_removed = str.charAt(0);
       
      // Remove first character
      return rem_str.substring(1,rem_str.length());
    }
 
    // If remaining string becomes
    // empty and last removed character
    // is same as first character of
    // original string. This is needed
    // for a string like "acbbcddc"
    if (rem_str.length() == 0 && last_removed ==
                                      str.charAt(0))
      return rem_str;
 
    // If the two first characters
    // of str and rem_str don't match, 
    // append first character of str
    // before the first character of
    // rem_str
    return (str.charAt(0) + rem_str);
  }
 
  static String remove(String str) 
  {
    char last_removed = '\0';
    return removeUtil(str, last_removed);      
  }
 
  // Driver code
  public static void main(String args[])
  {
    String str1 = "geeksforgeeg";
    System.out.println(remove(str1));
 
    String str2 = "azxxxzy";
    System.out.println(remove(str2));
 
    String str3 = "caaabbbaac";
    System.out.println(remove(str3));
 
    String str4 = "gghhg";
    System.out.println(remove(str4));
 
    String str5 = "aaaacddddcappp";
    System.out.println(remove(str5));
 
    String str6 = "aaaaaaaaaa";
    System.out.println(remove(str6));
 
    String str7 = "qpaaaaadaaaaadprq";
    System.out.println(remove(str7));
 
    String str8 = "acaaabbbacdddd";
    System.out.println(remove(str8));
  }
}
 
// This code is contibuted by rachana soma

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Python

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# Python program to remove
# all adjacent duplicates from a string
 
# Recursively removes adjacent
# duplicates from str and returns
# new string. las_removed is a
# pointer to last_removed character
def removeUtil(string, last_removed):
 
    # If length of string is 1 or 0
    if len(string) == 0 or len(string) == 1:
        return string
 
    # Remove leftmost same characters
    # and recur for remaining
    # string
    if string[0] == string[1]:
        last_removed = ord(string[0])
        while len(string) > 1 and string[0] ==
                                    string[1]:
            string = string[1:]
        string = string[1:]
 
        return removeUtil(string, last_removed)
 
    # At this point, the first
    # character is definiotely different
    # from its adjacent. Ignore first
    # character and recursively
    # remove characters from remaining string
    rem_str = removeUtil(string[1:], last_removed)
 
    # Check if the first character
    # of the rem_string matches
    # with the first character of
    # the original string
    if len(rem_str) != 0 and rem_str[0] ==
                                         string[0]:
        last_removed = ord(string[0])
        return (rem_str[1:])
 
    # If remaining string becomes
    # empty and last removed character
    # is same as first character of
    # original string. This is needed
    # for a string like "acbbcddc"
    if len(rem_str) == 0 and last_removed ==
                                   ord(string[0]):
        return rem_str
 
    # If the two first characters of
    # str and rem_str don't match,
    # append first character of str
    # before the first character of
    # rem_str.
    return ([string[0]] + rem_str)
 
def remove(string):
    last_removed = 0
    return toString(removeUtil(toList(string),
                                    last_removed))
 
# Utility functions
def toList(string):
    x = []
    for i in string:
        x.append(i)
    return x
 
def toString(x):
    return ''.join(x)
 
# Driver program
string1 = "geeksforgeeg"
print remove(string1)
 
string2 = "azxxxzy"
print remove(string2)
 
string3 = "caaabbbaac"
print remove(string3)
 
string4 = "gghhg"
print remove(string4)
 
string5 = "aaaacddddcappp"
print remove(string5)
 
string6 = "aaaaaaaaaa"
print remove(string6)
 
string7 = "qpaaaaadaaaaadprq"
print remove(string7)
 
string8 = "acaaabbbacdddd"
print remove(string8)
 
string9 = "acbbcddc"
print remove(string9)
 
# This code is contributed by BHAVYA JAIN

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Output: 

gksfor
ay

g
a

qrq
acac
a

Time Complexity: The time complexity of the solution can be written as T(n) = T(n-k) + O(k) where n is length of the input string and k is the number of first characters which are same. Solution of the recurrence is O(n)
Thanks to Prachi Bodke for suggesting this problem and initial solution. 

Another Approach:
The idea here is to check whether the String remStr has the repeated character that matches the last char of the parent String. If that is happening then we have to keep removing that character before concatenating string s and string remStr.
Below is the implementation of the above approach:

Java

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// Java Program for above approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
 
  // Recursively removes adjacent
  // duplicates from str and
  // returns new string. las_removed
  // is a pointer to
  // last_removed character
  private static String removeDuplicates(
    String s, char ch)
  {
 
    // If length of string is 1 or 0
    if (s == null || s.length() <= 1)
    {
      return s;
    }
 
    int i = 0;
    while (i < s.length())
    {
      if (i + 1 < s.length()
          && s.charAt(i) == s.charAt(i + 1))
      {
        int j = i;
        while (j + 1 < s.length()
               && s.charAt(j) ==
               s.charAt(j + 1))
        {
          j++;
        }
        char lastChar
          = i > 0 ? s.charAt(i - 1) : ch;
 
        String remStr = removeDuplicates(
          s.substring(j + 1, s.length()),
          lastChar);
 
        s = s.substring(0, i);
        int k = s.length(), l = 0;
 
        // Recursively remove all the adjacent
        // characters formed by removing the
        // adjacent characters
        while (remStr.length() > 0 &&
               s.length() > 0 &&
               remStr.charAt(0) ==
               s.charAt(s.length() - 1))
        {
 
          // Have to check whether this is the
          // repeated character that matches the
          // last char of the parent String
          while (remStr.length() > 0
                 && remStr.charAt(0) != ch
                 && remStr.charAt(0)
                 == s.charAt(s.length()
                             - 1))
          {
            remStr = remStr.substring(
              1, remStr.length());
          }
          s = s.substring(0, s.length() - 1);
        }
        s = s + remStr;
        i = j;
      }
      else
      {
        i++;
      }
    }
    return s;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    String str1 = "mississipie";
    System.out.println(removeDuplicates(
      str1, ' '));
    String str2 = "ocvvcolop";
    System.out.println(removeDuplicates(
      str2, ' '));
  }
}
 
// This code is contibuted by Niharika Sahai

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Output:

mpie
lop

Time Complexity: O(n) 
Thanks to Niharika Sahai for suggesting this problem and initial solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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